The pressure would double, yes, but it is confusing to write "the pressure is 2P". We need P to be a variable, standing for different pressures at different volumes. If the initial pressure and volume are P0 and V0 then when V=4V0 we will have P = 2P0.

So, in general for this question, if the initial pressure and volume are P0 and V0, what will P be when the volume is V?

The pressure would double, yes, but it is confusing to write "the pressure is 2P". We need P to be a variable, standing for different pressures at different volumes. If the initial pressure and volume are P0 and V0 then when V=4V0 we will have P = 2P0.

So, in general for this question, if the initial pressure and volume are P0 and V0, what will P be when the volume is V?

Staff: Mentor

The term "pressure rises as the square root of specific volume rises" means that the pressure is proportional to the square root of V (not equal to it). If, for an arbitrary state of the system, the pressure is P and the volume is V, what is the relationship between P and V (given P is proportional to the square root of V; you can use a proportionality constant k in your equation)?

The term "pressure rises as the square root of specific volume rises" means that the pressure is proportional to the square root of V (not equal to it). If, for an arbitrary state of the system, the pressure is P and the volume is V, what is the relationship between P and V (given P is proportional to the square root of V; you can use a proportionality constant k in your equation)?

No, k is the proportionality constant. It cannot be equal to the square root of itself (unless it is 1).
You wrote that the pressure would be equal to the square root of V. That cannot be right because V is a volume and P is a pressure, and the square root of a volume is not a pressure. The proportionality constant fixes that up. So instead of ##P=\sqrt V##, what would you write? You just need a factor k.